Question

y'+y=0 find a general solution for the homogeneous differential equations

Answer 1

^ is that linear?

Answer 2

Yes it is, because if y1 and y2 are solutions, then \(\lambda\, y_1+\mu\, y_2\) is a solution too.

Answer 3

It is linear so I used integration factor

Answer 4

\[\frac{ dy }{ dt }+a(t)y=0\]

Answer 5

\[e^{A(t)}; A(t)=\int\limits_{}^{}1 dt\]

Answer 6

so I got \[Ce^t\]

Answer 7

but the answer is \[Ce^{-t}\]

Answer 8

@phi

Answer 9

\[ \frac{ dy }{ dt }+y =0\]
multiply by the int. factor:
\[ e^t\frac{ dy }{ dt }+e^ty=0 \\

\frac{ d }{ dt }\left( e^ty\right) = 0 \\
e^ty = C\\
y = Ce^{-t}
\]

Answer 10

got you

Answer 11

@phi shouldn't the C be -C or it just don't matter?

Answer 12

C is an unknown constant. It could be plus or minus, but unless we are given more info, we don't know. -C is ok, but unless you have a good reason to toss in extra symbols, I would leave it just C)